$\mathcal{O}_K.\mathcal{O}_L=\mathcal{O}_{KL}$ The question is give a proof of counterexample. I know that I am looking for a counterexample, the statement is not true when the Disc(K) and the Disc(L) are not coprime. 
I have been trying to use $Disc(K)=[\mathcal{O}_K:K]^2Disc(\mathcal{O}_K)$, but am not having much success.
I have particularly been looking at the two fields generated by $f(x)=x^2-3$ and $g(x)=x^2-2$.
Any help would be greatly appreciated I have been trying to do this for ages. 
 A: If you're not particularly attached to your example, it's a little bit easier to avoid 2's as much as possible (since indeed, they are for biquadratic fields precisely the obstruction to $\mathcal{O}_K\mathcal{O}_L=\mathcal{O}_{KL}$ being true.)  There's important theory floating around in the background here, but since you just want an example for now:
Let $K=\mathbb{Q}(\sqrt{3})$ and $L=\mathbb{Q}(\sqrt{7})$.  Then since $3\equiv 7\equiv 3\pmod{4}$, a basis for the rings of integers are given respectively by $\mathcal{O}_K=\langle 1,\sqrt{3}\rangle$ and $\mathcal{O}_L=\langle 1,\sqrt{7}\rangle$ respectively.  So 
$$
\mathcal{O}_K\mathcal{O}_L=\{a+b\sqrt{3}+c\sqrt{7}+d\sqrt{21}\mid a,b,c,d\in\mathbb{Z}\},
$$
but it's easy to check that 
$$
\frac{\sqrt{3}+\sqrt{7}}{2}\in KL
$$
is an algebraic integer (indeed, its minimal polynomial is $x^4-5x^2+1$), so represents an element of $\mathcal{O}_{KL}$ not in $\mathcal{O}_{K}\mathcal{O}_{L}$.
Edit to add a couple of points:


*

*  I chose an unnecessarily sneaky example above because it's kind of non-obvious.  A somewhat easier approach is to note that $KL$ contains $\mathbb{Q}(\sqrt{21})$, and so since $21\equiv 1\pmod{4}$, we have $\frac{1+\sqrt{21}}{2}\in \mathcal{O}_{KL}$.

*  For the example you started with, $\sqrt{2}+\frac{1+\sqrt{3}}{2}$ works.

*  For a general analysis of the ring of integers of biquadratic fields, Kenneth Williams's "Integers of Biquadratic Fields" is extraordinarily clear and explicit.

